cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A322360 Multiplicative with a(p^e) = p^2 - 1.

Original entry on oeis.org

1, 3, 8, 3, 24, 24, 48, 3, 8, 72, 120, 24, 168, 144, 192, 3, 288, 24, 360, 72, 384, 360, 528, 24, 24, 504, 8, 144, 840, 576, 960, 3, 960, 864, 1152, 24, 1368, 1080, 1344, 72, 1680, 1152, 1848, 360, 192, 1584, 2208, 24, 48, 72, 2304, 504, 2808, 24, 2880, 144, 2880, 2520, 3480, 576, 3720, 2880, 384, 3, 4032, 2880
Offset: 1

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Author

Antti Karttunen, Dec 04 2018

Keywords

Comments

Absolute values of A046970, the Dirichlet inverse of the Jordan function J_2 (A007434).
Absolute values of the Möbius transform of A055491. (See Benoit Cloitre's May 31 2002 comment in A046970).

Links

Crossrefs

Absolute values of A046970.
Cf. A048250, A173557, A066086, A322359, A351265.

Programs

  • Maple
    a:= n-> mul(i[1]^2-1, i=ifactors(n)[2]):
seq(a(n), n=1..80);  # Alois P. Heinz, Jan 05 2021
  • Mathematica
    a[n_] := If[n==1, 1, Times @@ ((#^2-1)& @@@ FactorInteger[n])]; Array[a, 50]  (* Amiram Eldar, Dec 05 2018 *)
  • PARI
    A322360(n) = factorback(apply(p -> (p*p)-1, factor(n)[, 1]));
  • PARI
    A322360(n) = abs(sumdiv(n,d,moebius(n/d)*(core(d)^2)));

Formula

Multiplicative with a(p^e) = p^2 - 1.
a(n) = Product_{p prime divides n} (p^2 - 1).
a(n) = abs(A046970(n)).
a(n) = A048250(n) * A173557(n) = A066086(n) * A322359(n).
G.f. for a signed version of the sequence: Sum_{n >= 1} mu(n)*n^2*x^n/(1 - x^n) = Sum_{n >= 1} (-1)^omega(n)*a(n)*x^n = x - 3*x^2 - 8*x^3 - 3*x^4 - 24*x^5 + 24*x^6 - 48*x^7 - 3*x^8 - 8*x^9 + 72*x^10 - ..., where mu(n) is the Möbius function A008683(n) and omega(n) = A001221(n) is the number of distinct primes dividing n. - Peter Bala, Mar 05 2022
Sum_{k=1..n} a(k) ~ c * n^3, where c = (1/3) * Product_{p prime} (p-1)*(p^2 + 2*p + 2)/(p*(p^2 + p + 1)) = 0.187556464... . - Amiram Eldar, Oct 22 2022
a(n) = A007434(A007947(n)). - Enrique Pérez Herrero, Oct 14 2024

A374537 a(n) is the sum of the squares of the divisors of n that are exponentially odd numbers.

Original entry on oeis.org

1, 5, 10, 5, 26, 50, 50, 69, 10, 130, 122, 50, 170, 250, 260, 69, 290, 50, 362, 130, 500, 610, 530, 690, 26, 850, 739, 250, 842, 1300, 962, 1093, 1220, 1450, 1300, 50, 1370, 1810, 1700, 1794, 1682, 2500, 1850, 610, 260, 2650, 2210, 690, 50, 130, 2900, 850, 2810
Offset: 1

Views

Author

Amiram Eldar, Jul 11 2024

Keywords

Comments

The number of divisors of n that are exponentially odd is A322483(n) and their sum is A033634(n).

Links

Crossrefs

Cf. A001157, A005117, A033634, A322483, A351265, A358346, A374458, A374538.
Cf. A002117, A013661, A065464, A183699.

Programs

  • Mathematica
    f[p_, e_] := 1 + p^2 * (p^(4*Floor[(e-1)/2]+4) - 1) / (p^4 - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), p = f[, 1], e = f[, 2]); prod(i = 1, #p, 1 + p[i]^2 * (p[i]^(4*((e[i]-1)\2)+4) - 1) / (p[i]^4 - 1));}

Formula

a(n) = A001157(n) if and only if n is squarefree (A005117).
Multiplicative with a(p^e) = 1 + p^2 * (p^(4*floor((e-1)/2)+4) - 1) / (p^4 - 1).
Dirichlet g.f.: zeta(s) * zeta(2*s-4) * Product_{p prime} (1 + 1/p^(s-2) - 1/p^(2*s-4)).
Sum_{k=1..n} a(k) = c * n^3 / 3, where c = zeta(2) * zeta(3) * Product_{p prime} (1 - 2/p^2 + 1/p^3) = A183699 * A065464 = 0.84677961058798544766... .

A374539 The sum of the squares of the infinitary divisors of n.

Original entry on oeis.org

1, 5, 10, 17, 26, 50, 50, 85, 82, 130, 122, 170, 170, 250, 260, 257, 290, 410, 362, 442, 500, 610, 530, 850, 626, 850, 820, 850, 842, 1300, 962, 1285, 1220, 1450, 1300, 1394, 1370, 1810, 1700, 2210, 1682, 2500, 1850, 2074, 2132, 2650, 2210, 2570, 2402, 3130, 2900
Offset: 1

Views

Author

Amiram Eldar, Jul 11 2024

Keywords

Comments

Also the sum of the infinitary divisors of n^2.

Links

Crossrefs

Cf. A049417, A050376, A077609.
Similar sequences: A001157, A034676, A050999, A067558, A076577, A351265, A351307, A351647.

Programs

  • Mathematica
    f[p_, e_] := p^(2^Position[Reverse@IntegerDigits[e, 2], ?(# == 1 &)]); a[1] = 1; a[n] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n), b); prod(i = 1, #f~, b = binary(2 * f[i, 2]); prod(k=1, #b, if(b[k], 1+f[i, 1]^(2^(#b-k)), 1)));}
  • Python
    from math import prod
from sympy import factorint
def A374539(n): return prod(p**(1<Chai Wah Wu, Jul 11 2024

Formula

a(n) = A049417(n^2).
a(n) <= A001157(n), with equality if and only if n is in A036537.
Multiplicative with a(p^e) = Product{k>=1, e_k=1} (p^(2^(k+1)) + 1), where e = Sum_{k} e_k * 2^k is the binary representation of e, i.e., e_k is bit k of e.
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = Product_{P} (1 + 1/(P^2*(P+1))) = 1.14142906130350119631..., and P are numbers of the form p^(2^k) where p is prime and k >= 0 (A050376).
Previous Showing 11-13 of 13 results.

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